Math ! identify the vertex and the y-intercept of y=-2(x+3)^2+2the graph of the function . a. vertex: (−3, 2); y-intercept: −16 b. vertex: (−3, −2); y-intercept: 11 c. vertex: (3, 2); y-intercept: −16 d. vertex: (3, −2); y-intercept: −18 identify the vertex and the y-intercept of the graph of the function y=-2(x+2)^2+2 a. vertex: (−2, 2); y-intercept: −6 b. vertex: (2, 2); y-intercept: −6 c. vertex: (−2, −2); y-intercept: 6 d. vertex: (2, −2); y-intercept: −8

A and A

the equation of a parabola in vertex form is

y = a(x - h)² + k

where ( h, k ) are the coordinates of the vertex and a is a multiplier

y = - 2(x + 3)² + 2  is in this form

with vertex = ( - 3, 2)

To find the y-intercept let x = 0

y = - 2(3)² + 2 = - 18 + 2 = - 16

Similarly

y = - 2(x + 2)² + 2 is in vertex form

vertex = ( - 2 , 2)

x = 0 : y = - 2(2)² + 2 = - 8 + 2 = - 6 ← y- intercept

i would need to see the dot plot : )

just take the area of the square and then the area of the circle then subtract the circle area from the squares area

step-by-step explanation: